If ${\tan ^{ - 1}}\left( {{\rm{\alpha }} + i{\rm{\beta }}} \right) = x + iy$, then $x$ is equal to

$\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{2{\rm{\alpha }}}}{{1 - {{\rm{\alpha }}^2} - {{\rm{\beta }}^2}}}} \right)$

$\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{2{\rm{\alpha }}}}{{1 + {{\rm{\alpha }}^2} + {{\rm{\beta }}^2}}}} \right)$

${\tan ^{ - 1}}\left( {\frac{{2{\rm{\alpha }}}}{{1 - {{\rm{\alpha }}^2} - {{\rm{\beta }}^2}}}} \right)$

If $|z|=1$ and $z \neq 0,$ then the points $z+\frac{1}{z}$ lie on a

Straight line not through the origin

Line segment lying on the y-axis

If $z\bar z = 0$, iff

${\mathop{\rm Re}\nolimits} (z) = 0$

${\mathop{\rm Im}\nolimits} (z) = 0$

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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 181-210 of 979 questions

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181 Easy

181 of 979

Application +4 | -1

{z_1} = 1 + 2i

and

{z_2} = 3 + 5i

, then

{\mathop\mathrm{Re}\nolimits} \left[ {{{\bar z}_2}{z_1}/{z_2}} \right]

is equal to

View answer and explanation

Correct option: D

\frac{{22}}{{17}}

Option D (22/17) is correct because it is the real part of the complex number obtained after simplifying the given expression.

182 Easy

182 of 979

Application +4 | -1

The value of

\left(\frac{1+i}{\sqrt{2}}\right)^8+\left(\frac{1-i}{\sqrt{2}}\right)^8

is equal to

View answer and explanation

Correct option: D

Option D (2) is correct because after converting to polar form and applying De Moivre's theorem, the expression simplifies to 2.

183 Medium

183 of 979

Analyzing +4 | -1

a,\;b,\;c \ne 0

and belongs to the set

\left\{ {0,\;1,2,\;3,\; \ldots \ldots ..,\;9} \right\}

,
Then

{\log _{10}}\left( {\frac{{a + 10b + {{10}^2}c}}{{{{10}^{ - 4}}a + {{10}^{ - 3}}b + {{10}^{ - 2}}c}}} \right)

is equal to

View answer and explanation

Correct option: D

Option D (4) is correct because the expression simplifies to

{\log _{10}}{10^4}

, which equals 4.

184 Easy

184 of 979

Application +4 | -1

The value of

(1+i)^8+(1-i)^8

2^n

. Value of

n

View answer and explanation

Correct option: A

Option A (5) is correct because

(1+i)^8+(1-i)^8

simplifies to

2^5

, thus n=5.

185 Medium

185 of 979

Application +4 | -1

\tan ^{-1}(\alpha+i \beta)=x+i y

, then

x

is equal to

View answer and explanation

Correct option: B

\frac{1}{2} \tan ^{-1}\left(\frac{2 \alpha}{1+\alpha^2+\beta^2}\right)

Option A (

\frac{1}{2} \tan ^{-1}\left(\frac{2 \alpha}{1-\alpha^2-\beta^2}\right)

) is correct because using the tangent addition formula and taking the real part leads to the correct expression for

x

186 Easy

186 of 979

Application +4 | -1

The locus of the complex number

z

in the Argand plane if

\left|\frac{z-3}{z+3}\right|=2

, is

View answer and explanation

Correct option: C

A circle

Option C (A circle) is correct because the given condition represents a circle in the Argand plane.

187 Easy

187 of 979

Application +4 | -1

z = \frac{{\sqrt 3 }}{2} + \frac{i}{2}(i = \sqrt { - 1} ),

then

{(1 + iz + {z^5} + i{z^8})^9}

is eqaul to

View answer and explanation

Correct option: C

-1

Option C (-1) is correct because after simplifying the expression using Euler's formula and trigonometric identities, the result is -1.

188 Medium

188 of 979

Application +4 | -1

\left|z-\frac{4}{z}\right|=2

, then find the maximum value of

|z|

is equal to

View answer and explanation

Correct option: D

\sqrt{5}+1

The correct answer is D (

\sqrt{5}+1

) because it represents the maximum value of |z| derived from the given inequality using the triangle inequality.

189 Easy

189 of 979

Understanding +4 | -1

\arg \overline{{z}}+\arg {z} ; {z} \neq 0

is equal to:

View answer and explanation

Correct option: A

Option A (0) is correct because the argument of the conjugate of a complex number is the negative of the argument of the original complex number.

190 Medium

190 of 979

Application +4 | -1

{\tan ^{ - 1}}\left( {\mathrm{\alpha } + i\mathrm{\beta }} \right) = x + iy

, then

x

is equal to

View answer and explanation

Correct option: A

\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{2\mathrm{\alpha }}}{{1 - {\mathrm{\alpha }^2} - {\mathrm{\beta }^2}}}} \right)

Option A

\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{2\mathrm{\alpha }}}{{\left( {1 - {\mathrm{\alpha }^2} - {\mathrm{\beta }^2}} \right)}}} \right)

is the correct expression for x, derived by using the tangent addition formula and properties of complex conjugates.

191 Easy

191 of 979

Application +4 | -1

In the argand plane, all the complex numbers satisfying

|z-4 i|+|z+4 i|=10

lie on

View answer and explanation

Correct option: C

an ellipse

Option C (an ellipse) is correct because the given equation represents the locus of a point where the sum of its distances from two fixed points is constant, which is the definition of an ellipse.

192 Easy

192 of 979

Application +4 | -1

\frac{{{{\left( {\cos \theta + i\sin \theta } \right)}^4}}}{{{{\left( {\sin \theta + i\cos \theta } \right)}^5}}}

is equal to

View answer and explanation

Correct option: D

\sin 9\theta - i\cos 9\theta

Option D (

\sin 9\theta - i\cos 9\theta

) is correct because by applying De Moivre's theorem and simplifying, the expression reduces to the given form.

193 Easy

193 of 979

Analyzing +4 | -1

z\left( { \ne - 1} \right)

is a complex number such that

\frac{{z - 1}}{{z + 1}}

is purely imaginary, then

\left| z \right| =

View answer and explanation

Correct option: A

Option A (1) is correct because the given condition implies that the magnitude of z is 1.

194 Easy

194 of 979

Application +4 | -1

If the centre of a regular hexagon is at the origin and one of its vertices on argand diagram is

1 + 2i,

then its perimeter is

View answer and explanation

Correct option: D

6\sqrt 5

Option D (

6\sqrt 5

) is correct because the perimeter of a regular hexagon is six times the length of one of its sides, which is equal to the distance from the origin to a vertex.

195 Easy

195 of 979

Application +4 | -1

The value of

{\left[ {\sqrt 2 \left\{ {\cos \left( {{{56}^0}15'} \right) + i\sin \left( {{{56}^0}\;15'} \right)} \right\}} \right]^8}

View answer and explanation

Correct option: C

16i

Option C (

16i

) is correct because, using De-Moivre's Theorem, the expression simplifies to

16i

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196 Easy

196 of 979

Analyzing +4 | -1

a,\;b,\;c

are distinct positive numbers each being different from 1 such that

\left( {{{\log }_b}a.{{\log }_c}a - {{\log }_a}a} \right) + \left( {{{\log }_a}b.{{\log }_c}b - {{\log }_b}b} \right) + \left( {{{\log }_a}c.{{\log }_b}c - {{\log }_c}c} \right) = 0

, then

abc

View answer and explanation

Correct option: C

Option C (1) is correct because simplifying the given logarithmic expression leads to log(abc) = 0, which implies abc = 1.

197 Medium

197 of 979

Analyzing +4 | -1

z

is any complex number satisfying

|z-1|=1

then which of the following is correct?

View answer and explanation

Correct option: A

\arg (z-1)=2 \arg (z)

Option A (

\arg (z-1)=2 \arg (z)

) is correct because, by letting

z-1=e^{i \theta}

, we can express

z

in terms of

\theta

and show that

\arg(z) = \frac{{\theta}}{{2}}

, which leads to

\arg(z-1) = 2 \arg(z)

198 Easy

198 of 979

Application +4 | -1

The conjugate of a complex number is

\frac{1}{{i - 1}}

. Then, that complex number is

View answer and explanation

Correct option: D

- \frac{1}{{i + 1}}

Option D (

- \frac{1}{{i + 1}}

) is correct because it's the complex number whose conjugate is the given expression.

199 Medium

199 of 979

Application +4 | -1

Z_1=\frac{1}{a+i}, a \neq 0

and

Z_2=\frac{1}{1+\mathrm{bi}} \quad b \neq 0

such that

Z_1=\bar{Z}_2

then

View answer and explanation

Correct option: C

\mathrm{a}=1, \mathrm{~b}=-1

Option C (a=1, b=-1) is correct because it satisfies the condition Z1 = conjugate(Z2) by equating real and imaginary parts.

200 Easy

200 of 979

Application +4 | -1

x=a+b, y=a \omega+b \omega^2, z=a \omega^2+b \omega

, then

x y z

is equal to

View answer and explanation

Correct option: B

a^3+b^3

Option B (

a^3+b^3

) is correct because the expression simplifies to

a^3 + b^3

using the properties of cube roots of unity.

201 Easy

201 of 979

Application +4 | -1

\sqrt {x + iy} = \pm \left( {a + ib} \right),

then

\sqrt { - x - iy}

is equal to

View answer and explanation

Correct option: C

\pm \left( {b - ia} \right)

Option C (

\pm \left( {b - ia} \right)

) is correct because it is derived by equating the real and imaginary parts of the given equation and then finding the square root of the negative conjugate.

202 Medium

202 of 979

Application +4 | -1

|z|=1

and

z \neq 0,

then the points

z+\frac{1}{z}

lie on a

View answer and explanation

Correct option: C

Line segment lying on the x-axis

Option C is correct. Since |z|=1, z can be represented as e^(iθ). Therefore, z + 1/z = 2cos(θ), which is a real number between -2 and 2, representing a line segment on the x-axis.

203 Medium

203 of 979

Application +4 | -1

Let

a=\operatorname{Im}\left(\frac{1+z^{2}}{2 i z}\right)

, where

z

is any non-zero complex number.

The set

\mathrm{A}=\{a:|z|=1

and

z \neq \pm 1\}

is equal to:

View answer and explanation

Correct option: A

(-1,1)

Option A ((-1,1)) is correct because by substituting z = x + iy and |z| = 1 into the expression for 'a', we find that a = x, and since z ≠ ±1, x lies in the open interval (-1, 1).

204 Medium

204 of 979

Application +4 | -1

If the cube roots of unity are

1, \omega, \omega^{2}

then the roots of the equation

(x-1)^{3}+8=0

, are

View answer and explanation

Correct option: C

-1,1-2 \omega, 1-2 \omega^{2}

Option C (

-1, 1-2 \omega, 1-2 \omega^{2}

) is correct because it represents the roots of the given equation after simplifying and using the properties of cube roots of unity.

205 Easy

205 of 979

Analyzing +4 | -1

Let

{z_1}\;\mathrm{and}\;{z_2}

be the non-real roots of the equation

3{z^2} + 3z + b = 0.

If the origin together with the points represented by

{z_1}

and

{z_2}

form an equilateral triangle, then the value of

b

View answer and explanation

Correct option: A

Option A (1) is correct because the condition for the origin and the two roots to form an equilateral triangle leads to a specific relationship between the roots and the coefficient b.

206 Easy

206 of 979

Understanding +4 | -1

z\bar z = 0

, iff

View answer and explanation

Correct option: C

(z) = 0

Option C (

(z) = 0

) is correct because

z\bar z = |z|^2 = 0

implies that the magnitude of z is 0, which means z itself is 0.

207 Easy

207 of 979

Application +4 | -1

x=\cos \alpha+i \sin \alpha

and

y=\cos \beta+\mathrm{i} \sin \beta

, then

(x-y) /

(x+y)

is equal to

View answer and explanation

Correct option: A

i \tan \left(\frac{\alpha-\beta}{2}\right)

Option A (

i \tan \left( {\frac{{\alpha - \beta}}{2}} \right)

) is correct because using trigonometric identities, the expression simplifies to

i \tan \left( {\frac{{\alpha - \beta}}{2}} \right)

208 Medium

208 of 979

Application +4 | -1

\left( {{a_1} + i\;{b_1}} \right)\left( {{a_2} + i\;{b_2}} \right) \ldots \left( {{a_n} + i\;{b_n}} \right) = A + i\;B,

then

\left( {a_1^2 + b_1^2} \right)\left( {a_2^2 + b_2^2} \right) \ldots \left( {a_n^2 + b_n^2} \right)

is equal to

View answer and explanation

Correct option: B

{A^2} + {B^2}

Option B (

{A^2} + {B^2}

) is correct because the magnitude of the product of complex numbers is the product of their magnitudes.

209 Easy

209 of 979

Analyzing +4 | -1

Suppose

a \in \mathbf{R}

and the equation

z+a|z|+

2 i=0

has no solution in

\mathbf{C}

, then

a

satisfies the relation.

View answer and explanation

Correct option: B

|a| \geq 1

Option B (

|a| \geq 1

) is correct because the equation has no solution when the coefficient of |z|^2 is non-positive, which occurs when

|a| \geq 1

210 Medium

210 of 979

Application +4 | -1

z

satisfies the equation

|z|-z=1+2 i

, then

z

is equal to

View answer and explanation

Correct option: B

\frac{3}{2}-2 i

Option B (

\frac{3}{2}-2 i

) is correct because it satisfies the given equation

|z|-z=1+2i

when substituting

z=x+iy

Revision notes and key takeaways

Review COMPLEX NUMBERS in your textbook or class notes, then use this set to check your understanding.

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